TITLE: Prove that the intersection of all subfields of the reals is the rationals
QUESTION [12 upvotes]: I'm reading through Abstract Algebra by Hungerford and he makes the remark that the intersection of all subfields of the real numbers is the rational numbers.
Despite considerable deliberation, I'm unsure of the steps to take to show that the subfield is $\mathbb Q$. 
Any insight?

REPLY [19 votes]: First note that $\mathbb Q$ is itself a subfield of $\mathbb R$, so the intersection of all subfields must be a subset of the rationals.
Second note that $\mathbb Q$ is a prime field, that is, it has no proper subfields. This is true because if $F\subseteq\mathbb Q$ is a field then $1\in F$, deduce that $\mathbb N\subseteq F$, from this deduce that $\mathbb Z\subseteq F$ and then the conclusion.
Third, conclude the equality.

REPLY [6 votes]: Any subfield of the reals must contain 0 and 1. Since the subfield is closed under addition and subtraction, it must contain all the integers. Since it's also closed under division (except division by zero), it must contain the rationals.